\(\int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx\) [1488]

Optimal result

Integrand size = 27, antiderivative size = 189 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {\left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))}{16 d}+\frac {\left (15 a^2-48 a b+35 b^2\right ) \log (1+\sin (c+d x))}{16 d}-\frac {\left (a^2+3 b^2\right ) \sin (c+d x)}{d}-\frac {a b \sin ^2(c+d x)}{d}-\frac {b^2 \sin ^3(c+d x)}{3 d}-\frac {\sec ^2(c+d x) \left (20 a b+\left (9 a^2+11 b^2\right ) \sin (c+d x)\right )}{8 d}+\frac {\sec ^3(c+d x) (a+b \sin (c+d x))^2 \tan (c+d x)}{4 d} \] Output:

-1/16*(15*a^2+48*a*b+35*b^2)*ln(1-sin(d*x+c))/d+1/16*(15*a^2-48*a*b+35*b^2 
)*ln(1+sin(d*x+c))/d-(a^2+3*b^2)*sin(d*x+c)/d-a*b*sin(d*x+c)^2/d-1/3*b^2*s 
in(d*x+c)^3/d-1/8*sec(d*x+c)^2*(20*a*b+(9*a^2+11*b^2)*sin(d*x+c))/d+1/4*se 
c(d*x+c)^3*(a+b*sin(d*x+c))^2*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.84 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.98 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {-3 \left (15 a^2+48 a b+35 b^2\right ) \log (1-\sin (c+d x))+3 \left (15 a^2-48 a b+35 b^2\right ) \log (1+\sin (c+d x))+\frac {3 (a+b)^2}{(-1+\sin (c+d x))^2}+\frac {3 (a+b) (9 a+13 b)}{-1+\sin (c+d x)}-48 \left (a^2+3 b^2\right ) \sin (c+d x)-48 a b \sin ^2(c+d x)-16 b^2 \sin ^3(c+d x)-\frac {3 (a-b)^2}{(1+\sin (c+d x))^2}+\frac {3 (9 a-13 b) (a-b)}{1+\sin (c+d x)}}{48 d} \] Input:

Integrate[Sin[c + d*x]*(a + b*Sin[c + d*x])^2*Tan[c + d*x]^5,x]
 

Output:

(-3*(15*a^2 + 48*a*b + 35*b^2)*Log[1 - Sin[c + d*x]] + 3*(15*a^2 - 48*a*b 
+ 35*b^2)*Log[1 + Sin[c + d*x]] + (3*(a + b)^2)/(-1 + Sin[c + d*x])^2 + (3 
*(a + b)*(9*a + 13*b))/(-1 + Sin[c + d*x]) - 48*(a^2 + 3*b^2)*Sin[c + d*x] 
 - 48*a*b*Sin[c + d*x]^2 - 16*b^2*Sin[c + d*x]^3 - (3*(a - b)^2)/(1 + Sin[ 
c + d*x])^2 + (3*(9*a - 13*b)*(a - b))/(1 + Sin[c + d*x]))/(48*d)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3316, 27, 531, 25, 2176, 25, 2341, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sin (c+d x) \tan ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (c+d x)^6 (a+b \sin (c+d x))^2}{\cos (c+d x)^5}dx\)

\(\Big \downarrow \) 3316

\(\displaystyle \frac {b^5 \int \frac {\sin ^6(c+d x) (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {b^6 \sin ^6(c+d x) (a+b \sin (c+d x))^2}{\left (b^2-b^2 \sin ^2(c+d x)\right )^3}d(b \sin (c+d x))}{b d}\)

\(\Big \downarrow \) 531

\(\displaystyle \frac {\frac {\int -\frac {(a+b \sin (c+d x)) \left (4 \sin ^5(c+d x) b^7+4 \sin ^3(c+d x) b^7+3 \sin (c+d x) b^7+4 a \sin ^4(c+d x) b^6+4 a \sin ^2(c+d x) b^6+a b^6\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}+\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}}{b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\int \frac {(a+b \sin (c+d x)) \left (4 \sin ^5(c+d x) b^7+4 \sin ^3(c+d x) b^7+3 \sin (c+d x) b^7+4 a \sin ^4(c+d x) b^6+4 a \sin ^2(c+d x) b^6+a b^6\right )}{\left (b^2-b^2 \sin ^2(c+d x)\right )^2}d(b \sin (c+d x))}{4 b^2}}{b d}\)

\(\Big \downarrow \) 2176

\(\displaystyle \frac {\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {\int -\frac {8 \sin ^4(c+d x) b^8+16 a \sin ^3(c+d x) b^7+32 a \sin (c+d x) b^7+8 \left (a^2+2 b^2\right ) \sin ^2(c+d x) b^6+\left (7 a^2+11 b^2\right ) b^6}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}+\frac {b^4 (a+b \sin (c+d x)) \left (9 a b \sin (c+d x)+11 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}}{4 b^2}}{b d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x)) \left (9 a b \sin (c+d x)+11 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \frac {8 \sin ^4(c+d x) b^8+16 a \sin ^3(c+d x) b^7+32 a \sin (c+d x) b^7+8 \left (a^2+2 b^2\right ) \sin ^2(c+d x) b^6+\left (7 a^2+11 b^2\right ) b^6}{b^2-b^2 \sin ^2(c+d x)}d(b \sin (c+d x))}{2 b^2}}{4 b^2}}{b d}\)

\(\Big \downarrow \) 2341

\(\displaystyle \frac {\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x)) \left (9 a b \sin (c+d x)+11 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {\int \left (-8 \sin ^2(c+d x) b^6-16 a \sin (c+d x) b^5-8 \left (a^2+3 b^2\right ) b^4+\frac {48 a \sin (c+d x) b^7+5 \left (3 a^2+7 b^2\right ) b^6}{b^2-b^2 \sin ^2(c+d x)}\right )d(b \sin (c+d x))}{2 b^2}}{4 b^2}}{b d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {b^5 \sin (c+d x) (a+b \sin (c+d x))^2}{4 \left (b^2-b^2 \sin ^2(c+d x)\right )^2}-\frac {\frac {b^4 (a+b \sin (c+d x)) \left (9 a b \sin (c+d x)+11 b^2\right )}{2 \left (b^2-b^2 \sin ^2(c+d x)\right )}-\frac {5 b^5 \left (3 a^2+7 b^2\right ) \text {arctanh}(\sin (c+d x))-8 b^5 \left (a^2+3 b^2\right ) \sin (c+d x)-8 a b^6 \sin ^2(c+d x)-24 a b^6 \log \left (b^2-b^2 \sin ^2(c+d x)\right )-\frac {8}{3} b^7 \sin ^3(c+d x)}{2 b^2}}{4 b^2}}{b d}\)

Input:

Int[Sin[c + d*x]*(a + b*Sin[c + d*x])^2*Tan[c + d*x]^5,x]
 

Output:

((b^5*Sin[c + d*x]*(a + b*Sin[c + d*x])^2)/(4*(b^2 - b^2*Sin[c + d*x]^2)^2 
) - ((b^4*(a + b*Sin[c + d*x])*(11*b^2 + 9*a*b*Sin[c + d*x]))/(2*(b^2 - b^ 
2*Sin[c + d*x]^2)) - (5*b^5*(3*a^2 + 7*b^2)*ArcTanh[Sin[c + d*x]] - 24*a*b 
^6*Log[b^2 - b^2*Sin[c + d*x]^2] - 8*b^5*(a^2 + 3*b^2)*Sin[c + d*x] - 8*a* 
b^6*Sin[c + d*x]^2 - (8*b^7*Sin[c + d*x]^3)/3)/(2*b^2))/(4*b^2))/(b*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 531

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi 
alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a 
 + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 
2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b 
*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p 
 + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt 
Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2176

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a + b*x^2, x], R = Coeff[PolynomialRema 
inder[Pq, a + b*x^2, x], x, 0], S = Coeff[PolynomialRemainder[Pq, a + b*x^2 
, x], x, 1]}, Simp[(d + e*x)^m*(a + b*x^2)^(p + 1)*((a*S - b*R*x)/(2*a*b*(p 
 + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^(p 
 + 1)*ExpandToSum[2*a*b*(p + 1)*(d + e*x)*Qx - a*e*S*m + b*d*R*(2*p + 3) + 
b*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && R 
ationalQ[a, b, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

rule 2341

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* 
(a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 5.90 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.41

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{9}}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \sin \left (d x +c \right )^{7}}{8}-\frac {7 \sin \left (d x +c \right )^{5}}{8}-\frac {35 \sin \left (d x +c \right )^{3}}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(266\)
default	\(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{9}}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \sin \left (d x +c \right )^{7}}{8}-\frac {7 \sin \left (d x +c \right )^{5}}{8}-\frac {35 \sin \left (d x +c \right )^{3}}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(266\)
parts	\(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \sin \left (d x +c \right )^{7}}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \sin \left (d x +c \right )^{5}}{8}-\frac {5 \sin \left (d x +c \right )^{3}}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {b^{2} \left (\frac {\sin \left (d x +c \right )^{9}}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \sin \left (d x +c \right )^{9}}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \sin \left (d x +c \right )^{7}}{8}-\frac {7 \sin \left (d x +c \right )^{5}}{8}-\frac {35 \sin \left (d x +c \right )^{3}}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {2 a b \left (\frac {\sin \left (d x +c \right )^{8}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{8}}{2 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{6}}{2}-\frac {3 \sin \left (d x +c \right )^{4}}{4}-\frac {3 \sin \left (d x +c \right )^{2}}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(271\)
risch	\(\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i b^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {a b \,{\mathrm e}^{2 i \left (d x +c \right )}}{4 d}-\frac {i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {13 i {\mathrm e}^{i \left (d x +c \right )} b^{2}}{8 d}+\frac {12 i a b c}{d}-\frac {13 i {\mathrm e}^{-i \left (d x +c \right )} b^{2}}{8 d}+\frac {a b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{4 d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (9 a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+13 b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+5 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+48 i a b \,{\mathrm e}^{5 i \left (d x +c \right )}-a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-5 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+64 i a b \,{\mathrm e}^{3 i \left (d x +c \right )}-9 a^{2}-13 b^{2}+48 i a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {i b^{2} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+6 i x a b -\frac {15 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{d}-\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 d}+\frac {15 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}-\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{d}+\frac {35 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 d}\)	\(475\)

Input:

int(sin(d*x+c)*(a+b*sin(d*x+c))^2*tan(d*x+c)^5,x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^2*(1/4*sin(d*x+c)^7/cos(d*x+c)^4-3/8*sin(d*x+c)^7/cos(d*x+c)^2-3/8* 
sin(d*x+c)^5-5/8*sin(d*x+c)^3-15/8*sin(d*x+c)+15/8*ln(sec(d*x+c)+tan(d*x+c 
)))+2*a*b*(1/4*sin(d*x+c)^8/cos(d*x+c)^4-1/2*sin(d*x+c)^8/cos(d*x+c)^2-1/2 
*sin(d*x+c)^6-3/4*sin(d*x+c)^4-3/2*sin(d*x+c)^2-3*ln(cos(d*x+c)))+b^2*(1/4 
*sin(d*x+c)^9/cos(d*x+c)^4-5/8*sin(d*x+c)^9/cos(d*x+c)^2-5/8*sin(d*x+c)^7- 
7/8*sin(d*x+c)^5-35/24*sin(d*x+c)^3-35/8*sin(d*x+c)+35/8*ln(sec(d*x+c)+tan 
(d*x+c))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.05 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {48 \, a b \cos \left (d x + c\right )^{6} - 24 \, a b \cos \left (d x + c\right )^{4} + 3 \, {\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 144 \, a b \cos \left (d x + c\right )^{2} + 24 \, a b + 2 \, {\left (8 \, b^{2} \cos \left (d x + c\right )^{6} - 8 \, {\left (3 \, a^{2} + 10 \, b^{2}\right )} \cos \left (d x + c\right )^{4} - 3 \, {\left (9 \, a^{2} + 13 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, a^{2} + 6 \, b^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:

integrate(sin(d*x+c)*(a+b*sin(d*x+c))^2*tan(d*x+c)^5,x, algorithm="fricas" 
)
 

Output:

1/48*(48*a*b*cos(d*x + c)^6 - 24*a*b*cos(d*x + c)^4 + 3*(15*a^2 - 48*a*b + 
 35*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(15*a^2 + 48*a*b + 35*b^ 
2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) - 144*a*b*cos(d*x + c)^2 + 24*a*b 
 + 2*(8*b^2*cos(d*x + c)^6 - 8*(3*a^2 + 10*b^2)*cos(d*x + c)^4 - 3*(9*a^2 
+ 13*b^2)*cos(d*x + c)^2 + 6*a^2 + 6*b^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
 

Sympy [F]

\[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sin {\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx \] Input:

integrate(sin(d*x+c)*(a+b*sin(d*x+c))**2*tan(d*x+c)**5,x)
 

Output:

Integral((a + b*sin(c + d*x))**2*sin(c + d*x)*tan(c + d*x)**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.95 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=-\frac {16 \, b^{2} \sin \left (d x + c\right )^{3} + 48 \, a b \sin \left (d x + c\right )^{2} - 3 \, {\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 48 \, {\left (a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right ) - \frac {6 \, {\left (24 \, a b \sin \left (d x + c\right )^{2} + {\left (9 \, a^{2} + 13 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 20 \, a b - {\left (7 \, a^{2} + 11 \, b^{2}\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{48 \, d} \] Input:

integrate(sin(d*x+c)*(a+b*sin(d*x+c))^2*tan(d*x+c)^5,x, algorithm="maxima" 
)
 

Output:

-1/48*(16*b^2*sin(d*x + c)^3 + 48*a*b*sin(d*x + c)^2 - 3*(15*a^2 - 48*a*b 
+ 35*b^2)*log(sin(d*x + c) + 1) + 3*(15*a^2 + 48*a*b + 35*b^2)*log(sin(d*x 
 + c) - 1) + 48*(a^2 + 3*b^2)*sin(d*x + c) - 6*(24*a*b*sin(d*x + c)^2 + (9 
*a^2 + 13*b^2)*sin(d*x + c)^3 - 20*a*b - (7*a^2 + 11*b^2)*sin(d*x + c))/(s 
in(d*x + c)^4 - 2*sin(d*x + c)^2 + 1))/d
 

Giac [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.09 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {{\left (15 \, a^{2} - 48 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{16 \, d} - \frac {{\left (15 \, a^{2} + 48 \, a b + 35 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{16 \, d} - \frac {b^{2} d^{2} \sin \left (d x + c\right )^{3} + 3 \, a b d^{2} \sin \left (d x + c\right )^{2} + 3 \, a^{2} d^{2} \sin \left (d x + c\right ) + 9 \, b^{2} d^{2} \sin \left (d x + c\right )}{3 \, d^{3}} + \frac {24 \, a b \sin \left (d x + c\right )^{2} + {\left (9 \, a^{2} + 13 \, b^{2}\right )} \sin \left (d x + c\right )^{3} - 20 \, a b - {\left (7 \, a^{2} + 11 \, b^{2}\right )} \sin \left (d x + c\right )}{8 \, d {\left (\sin \left (d x + c\right ) + 1\right )}^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} \] Input:

integrate(sin(d*x+c)*(a+b*sin(d*x+c))^2*tan(d*x+c)^5,x, algorithm="giac")
 

Output:

1/16*(15*a^2 - 48*a*b + 35*b^2)*log(abs(sin(d*x + c) + 1))/d - 1/16*(15*a^ 
2 + 48*a*b + 35*b^2)*log(abs(sin(d*x + c) - 1))/d - 1/3*(b^2*d^2*sin(d*x + 
 c)^3 + 3*a*b*d^2*sin(d*x + c)^2 + 3*a^2*d^2*sin(d*x + c) + 9*b^2*d^2*sin( 
d*x + c))/d^3 + 1/8*(24*a*b*sin(d*x + c)^2 + (9*a^2 + 13*b^2)*sin(d*x + c) 
^3 - 20*a*b - (7*a^2 + 11*b^2)*sin(d*x + c))/(d*(sin(d*x + c) + 1)^2*(sin( 
d*x + c) - 1)^2)
 

Mupad [B] (verification not implemented)

Time = 20.34 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.29 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (\frac {15\,a^2}{8}-6\,a\,b+\frac {35\,b^2}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {15\,a^2}{8}+6\,a\,b+\frac {35\,b^2}{8}\right )}{d}-\frac {\left (-\frac {15\,a^2}{4}-\frac {35\,b^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}-12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+\left (\frac {5\,a^2}{2}+\frac {35\,b^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+\left (\frac {47\,a^2}{4}+\frac {329\,b^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+32\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\left (11\,a^2-17\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+32\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {47\,a^2}{4}+\frac {329\,b^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {5\,a^2}{2}+\frac {35\,b^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-12\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (-\frac {15\,a^2}{4}-\frac {35\,b^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}+\frac {6\,a\,b\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \] Input:

int(sin(c + d*x)*tan(c + d*x)^5*(a + b*sin(c + d*x))^2,x)
 

Output:

(log(tan(c/2 + (d*x)/2) + 1)*((15*a^2)/8 - 6*a*b + (35*b^2)/8))/d - (log(t 
an(c/2 + (d*x)/2) - 1)*(6*a*b + (15*a^2)/8 + (35*b^2)/8))/d - (tan(c/2 + ( 
d*x)/2)^7*(11*a^2 - 17*b^2) + tan(c/2 + (d*x)/2)^3*((5*a^2)/2 + (35*b^2)/6 
) + tan(c/2 + (d*x)/2)^11*((5*a^2)/2 + (35*b^2)/6) - tan(c/2 + (d*x)/2)^13 
*((15*a^2)/4 + (35*b^2)/4) + tan(c/2 + (d*x)/2)^5*((47*a^2)/4 + (329*b^2)/ 
12) + tan(c/2 + (d*x)/2)^9*((47*a^2)/4 + (329*b^2)/12) - tan(c/2 + (d*x)/2 
)*((15*a^2)/4 + (35*b^2)/4) - 12*a*b*tan(c/2 + (d*x)/2)^2 + 12*a*b*tan(c/2 
 + (d*x)/2)^4 + 32*a*b*tan(c/2 + (d*x)/2)^6 + 32*a*b*tan(c/2 + (d*x)/2)^8 
+ 12*a*b*tan(c/2 + (d*x)/2)^10 - 12*a*b*tan(c/2 + (d*x)/2)^12)/(d*(tan(c/2 
 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 - 3*tan(c/2 + (d*x)/2)^6 - 3*tan(c/ 
2 + (d*x)/2)^8 + 3*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 - tan(c/2 
 + (d*x)/2)^14 - 1)) + (6*a*b*log(tan(c/2 + (d*x)/2)^2 + 1))/d
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 617, normalized size of antiderivative = 3.26 \[ \int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^5(c+d x) \, dx =\text {Too large to display} \] Input:

int(sin(d*x+c)*(a+b*sin(d*x+c))^2*tan(d*x+c)^5,x)
 

Output:

(144*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**4*a*b - 288*log(tan((c + d 
*x)/2)**2 + 1)*sin(c + d*x)**2*a*b + 144*log(tan((c + d*x)/2)**2 + 1)*a*b 
- 45*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2 - 144*log(tan((c + d*x 
)/2) - 1)*sin(c + d*x)**4*a*b - 105*log(tan((c + d*x)/2) - 1)*sin(c + d*x) 
**4*b**2 + 90*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2 + 288*log(tan 
((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b + 210*log(tan((c + d*x)/2) - 1)*sin 
(c + d*x)**2*b**2 - 45*log(tan((c + d*x)/2) - 1)*a**2 - 144*log(tan((c + d 
*x)/2) - 1)*a*b - 105*log(tan((c + d*x)/2) - 1)*b**2 + 45*log(tan((c + d*x 
)/2) + 1)*sin(c + d*x)**4*a**2 - 144*log(tan((c + d*x)/2) + 1)*sin(c + d*x 
)**4*a*b + 105*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**2 - 90*log(tan 
((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2 + 288*log(tan((c + d*x)/2) + 1)*si 
n(c + d*x)**2*a*b - 210*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**2 + 4 
5*log(tan((c + d*x)/2) + 1)*a**2 - 144*log(tan((c + d*x)/2) + 1)*a*b + 105 
*log(tan((c + d*x)/2) + 1)*b**2 - 8*sin(c + d*x)**7*b**2 - 24*sin(c + d*x) 
**6*a*b - 24*sin(c + d*x)**5*a**2 - 56*sin(c + d*x)**5*b**2 + 108*sin(c + 
d*x)**4*a*b + 75*sin(c + d*x)**3*a**2 + 175*sin(c + d*x)**3*b**2 - 72*sin( 
c + d*x)**2*a*b - 45*sin(c + d*x)*a**2 - 105*sin(c + d*x)*b**2)/(24*d*(sin 
(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))